Question

identify the coordinates of any local and absolute extreme points and inflection points. graph the function.

( y = \frac { 8 x } { x ^ { 2 } + 9 } )

separate answers as needed.)

b. there are no inflection points.

now sketch the graph. choose the correct graph below.

a.
b.
c.
d.

Explanation:

Step1: Find the first derivative

Use the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = 8x\), \(u^\prime=8\), \(v=x^{2}+9\), \(v^\prime = 2x\).

$$
LATEXBLOCK0
$$

Set \(y^\prime = 0\), then \(72-8x^{2}=0\), \(x^{2}=9\), \(x=\pm3\).
When \(x = 3\), \(y=\frac{8\times3}{3^{2}+9}=\frac{24}{18}=\frac{4}{3}\).
When \(x=- 3\), \(y=\frac{8\times(-3)}{(-3)^{2}+9}=\frac{-24}{18}=-\frac{4}{3}\).
Since \(\lim_{x
ightarrow\pm\infty}\frac{8x}{x^{2}+9}=0\), the local maximum is \((3,\frac{4}{3})\) and local minimum is \((-3,-\frac{4}{3})\), and they are also absolute maximum and minimum.

Step2: Find the second derivative

Use the quotient rule again. Let \(u = 72 - 8x^{2}\), \(u^\prime=-16x\), \(v=(x^{2}+9)^{2}\), \(v^\prime = 2(x^{2}+9)\times2x = 4x(x^{2}+9)\)

$$
LATEXBLOCK1
$$

Set \(y^{\prime\prime}=0\), then \(16x(x^{2}-27)=0\), \(x = 0\) or \(x=\pm3\sqrt{3}\).
When \(x = 0\), \(y = 0\); when \(x=3\sqrt{3}\), \(y=\frac{8\times3\sqrt{3}}{(3\sqrt{3})^{2}+9}=\frac{24\sqrt{3}}{27 + 9}=\frac{\sqrt{3}}{2}\); when \(x=-3\sqrt{3}\), \(y=-\frac{\sqrt{3}}{2}\).
So the inflection points are \((0,0)\), \((3\sqrt{3},\frac{\sqrt{3}}{2})\), \((-3\sqrt{3},-\frac{\sqrt{3}}{2})\)

Answer:

Local and absolute maximum at \((3,\frac{4}{3})\), local and absolute minimum at \((-3,-\frac{4}{3})\), inflection points at \((0,0)\), \((3\sqrt{3},\frac{\sqrt{3}}{2})\), \((-3\sqrt{3},-\frac{\sqrt{3}}{2})\)